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LLL reduced

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Definition

Let δ∈(14,1)\delta\in\left(\frac14,1\right)δ∈(41​,1). A basis{bi}i=1d\left\{b_i\right\}_{i=1}^d{bi​}i=1d​is Ξ΄\deltaΞ΄- LLL-reduced if it is size reduced and satisfy the LovΓ‘sz condition, i.e.

Ξ΄βˆ₯biβˆ—βˆ₯2≀βˆ₯bi+1βˆ—+ΞΌi+1,ibiβˆ—βˆ₯2\delta\left\lVert b_i^*\right\rVert^2\leq\left\lVert b_{i+1}^*+\mu_{i+1,i}b_i^*\right\rVert^2Ξ΄βˆ₯biβˆ—β€‹βˆ₯2≀​bi+1βˆ—β€‹+ΞΌi+1,i​biβˆ—β€‹β€‹2

This notion of reduction is most useful to use for fast algorithms as such a basis can be found in polynomial time (see LLL reduction).

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Bounds

βˆ₯b1βˆ₯≀(44Ξ΄βˆ’1)dβˆ’14vol(L)1dβˆ₯biβˆ₯≀(44Ξ΄βˆ’1)dβˆ’12Ξ»i(L)∏i=1dβˆ₯biβˆ₯≀(44Ξ΄βˆ’1)d(dβˆ’1)4vol(L)\begin{align*} \left\lVert b_1\right\rVert&\leq\left(\frac4{4\delta-1}\right)^{\frac{d-1}4}\text{vol}(L)^\frac1d\\ \left\lVert b_i\right\rVert&\leq\left(\frac4{4\delta-1}\right)^{\frac{d-1}2}\lambda_i(L)\\ \prod_{i=1}^d\left\lVert b_i\right\rVert&\leq\left(\frac4{4\delta-1}\right)^{\frac{d(d-1)}4}\text{vol}(L) \end{align*}βˆ₯b1​βˆ₯βˆ₯bi​βˆ₯i=1∏d​βˆ₯bi​βˆ₯​≀(4Ξ΄βˆ’14​)4dβˆ’1​vol(L)d1​≀(4Ξ΄βˆ’14​)2dβˆ’1​λi​(L)≀(4Ξ΄βˆ’14​)4d(dβˆ’1)​vol(L)​

Lattice reduction

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Overview

Having introduced the LLL reduction, we now provide a more general notions of a reduced basis for a lattice as well as provide bounds for the basis vectors. The key idea behind introducing these definitions is that once we know some basis vector is []-reduced, we can bound the sizes of the basis, which is important when algorithms require short vectors in a lattice. For fast algorithms, LLL-reduction is typically the most important notion as it can be computed quickly. Two main definitions appear often when discussing lattice reductions, which we will provide here.

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Definitions

A basisis size-reduced if . Intuitively this captures the idea that a reduced basis being "almost orthogonal".

Let be a lattice, , we define the successive minima as

Intuitively, is the length of the " shortest lattice vector". This intuition is illustrated by the definition of :

However this is not precise as if is the shortest lattice vector, then is also the shortest lattice vector.

Unfortunately, a basisfor where for dimensions and above. This tells us that we can't actually define "the most reduced basis" in contrast to the 2D case (see ) and we would need some other definition to convey this intuition.

An alternate definition ofthat will be helpful is the radius of the smallest ball centered at the origin such that the ball contains at leastlinearly independent vectors in.

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Exercises

1) Show that both definitions of are equivalent

2) Consider the lattice . Show that the successive minima are allbut no basiscan satisfy .

{bi}i=1d\left\{b_i\right\}_{i=1}^d{bi​}i=1d​
∣μi,jβˆ£β‰€12\left|\mu_{i,j}\right|\leq\frac12∣μi,jβ€‹βˆ£β‰€21​
LLL
1≀i≀dim⁑L=d1\leq i\leq\dim L=d1≀i≀dimL=d
ithi^\text{th}ith
Ξ»i(L)\lambda_i(L)Ξ»i​(L)
Ξ»i(L)=min⁑(max⁑1≀j≀i(βˆ₯vjβˆ₯):vj∈LΒ areΒ linearlyΒ independent)\lambda_i(L)=\min\left(\max_{1\leq j\leq i}\left(\left\lVert v_j\right\rVert\right):v_j\in L\text{ are linearly independent}\right)Ξ»i​(L)=min(1≀j≀imax​(βˆ₯vj​βˆ₯):vjβ€‹βˆˆLΒ areΒ linearlyΒ independent)
Ξ»i(L)\lambda_i(L)Ξ»i​(L)
ithi^\text{th}ith
Ξ»1\lambda_1Ξ»1​
Ξ»1(L)=min⁑(βˆ₯vβˆ₯:v∈L)\lambda_1(L)=\min\left(\left\lVert v\right\rVert:v\in L\right)Ξ»1​(L)=min(βˆ₯vβˆ₯:v∈L)
vvv
βˆ’v-vβˆ’v
bib_ibi​
LLL
Ξ»i(L)=βˆ₯biβˆ₯\lambda_i(L)=\left\lVert b_i\right\rVertΞ»i​(L)=βˆ₯bi​βˆ₯
555
Ξ»i(L)\lambda_i(L)Ξ»i​(L)
iii
LLL
Ξ»i\lambda_iΞ»i​
L=(2000002000002000002011111)L=\begin{pmatrix}2&0&0&0&0\\0&2&0&0&0\\0&0&2&0&0\\0&0&0&2&0\\1&1&1&1&1\end{pmatrix}L=​20001​02001​00201​00021​00001​​
222
bib_ibi​
βˆ₯biβˆ₯=Ξ»i\left\lVert b_i\right\rVert=\lambda_iβˆ₯bi​βˆ₯=Ξ»i​
Lagrange's algorithm

HKZ reduced

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Definition

Let Ο€i\pi_iΟ€i​as the projection to the orthogonal complement of {bj}j=1iβˆ’1\left\{b_j\right\}_{j=1}^{i-1}{bj​}j=1iβˆ’1​.Then the basis is HKZ-reduced if it is size-reduced and ∣∣biβˆ—βˆ£βˆ£=Ξ»1(Ο€i(L))||b_i^*||=\lambda_1\left(\pi_i(L)\right)∣∣biβˆ—β€‹βˆ£βˆ£=Ξ»1​(Ο€i​(L)). This definition gives us a relatively simple way to compute a HKZ-reduced basis by iteratively finding the shortest vector in orthogonal projections.

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Bounds

4i+3≀(∣∣bi∣∣λi(L))2≀i+34\frac4{i+3}\leq\left(\frac{||b_i||}{\lambda_i(L)}\right)^2\leq\frac{i+3}4i+34​≀(Ξ»i​(L)∣∣biβ€‹βˆ£βˆ£β€‹)2≀4i+3​

Minkowski reduced

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Definition

The basis{bi}i=1d\left\{b_i\right\}_{i=1}^d{bi​}i=1d​ is Minkowski-reduced if bib_ibi​has minimum length among all vectors in LLL linearly independent from{bj}j=1iβˆ’1\left\{b_j\right\}_{j=1}^{i-1}{bj​}j=1iβˆ’1​. Equivalently, bib_ibi​has minimum length among all vectors vvvsuch that {b1,…,biβˆ’1,v}\left\{b_1,\dots,b_{i-1},v\right\}{b1​,…,biβˆ’1​,v}can be extended to form a basis ofLLL. Such a notion is strongest among all lattice reduction notions and is generally extremely hard to compute. Another equivalent definition is

βˆ₯biβˆ₯≀βˆ₯βˆ‘j=idcjbjβˆ₯gcd⁑(cj)=1\left\lVert b_i\right\rVert\leq\left\lVert\sum_{j=i}^dc_jb_j\right\rVert\quad\gcd\left(c_j\right)=1βˆ₯bi​βˆ₯≀​j=iβˆ‘d​cj​bj​​gcd(cj​)=1

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Bounds

The proof presented here is based off [Waerden 1956]. We proceed by induction. Letbe a Minkowski-reduced basis for some lattice. The lower bound is immediate and for , the upper bound is immediate as well.

Let be linearly independent vectors such that. Let be the sublattice generated by . Evidently somemust exist such thatis not in . Consider the new lattice . Letbe the shortest vector insuch thatis a basis for and we have

If, then we are done sincecan be extended to a basis of , so . Otherwise, we have . Let whereis the projection ofin. Since by definition we have, we must have

Furthermore, since

we have , hence we have

but since by definition, the case of cannot occur here (hence in these cases).

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Exercises

1) Show that both definitions of Minkowski-reduced lattice are equivalent

2) Consider the lattice . We have showed in a previous exercise that the successive minima are allbut no basiscan satisfy , show that for any Minkowski reduced basis , the basis must satisfy

Ξ»i(L)2≀βˆ₯biβˆ₯2≀max⁑(1,(54)iβˆ’4)Ξ»i(L)2\lambda_i(L)^2\leq\left\lVert b_i\right\rVert^2\leq\max\left(1,\left(\frac54\right)^{i-4}\right)\lambda_i(L)^2Ξ»i​(L)2≀βˆ₯bi​βˆ₯2≀max(1,(45​)iβˆ’4)Ξ»i​(L)2
bib_ibi​
LLL
i=1i=1i=1
v1,v2…viv_1,v_2\dots v_iv1​,v2​…vi​
βˆ₯vjβˆ₯=Ξ»j(L)\left\lVert v_j\right\rVert=\lambda_j(L)βˆ₯vj​βˆ₯=Ξ»j​(L)
Liβˆ’1L_{i-1}Liβˆ’1​
b1,b2,…biβˆ’1b_1,b_2,\dots b_{i-1}b1​,b2​,…biβˆ’1​
kkk
vkv_kvk​
Liβˆ’1L_{i-1}Liβˆ’1​
Lβ€²=L∩span(b1,b2,…biβˆ’1,vk)L'=L\cap\text{span}\left(b_1,b_2,\dots b_{i-1},v_k\right)Lβ€²=L∩span(b1​,b2​,…biβˆ’1​,vk​)
vkβ€²v'_kvk′​
Lβ€²βˆ’Liβˆ’1L'-L_{i-1}Lβ€²βˆ’Liβˆ’1​
b1,b2,…,biβˆ’1,vkβ€²b_1,b_2,\dots,b_{i-1},v'_kb1​,b2​,…,biβˆ’1​,vk′​
Lβ€²L'Lβ€²
vk=a1b1+a2b2+β‹―+aiβˆ’1biβˆ’1+nvkβ€²ai,n∈Zβˆ₯biβˆ₯≀βˆ₯vkβ€²βˆ₯v_k=a_1b_1+a_2b_2+\dots+a_{i-1}b_{i-1}+nv'_k\quad a_i,n\in\mathbb Z\\ \left\lVert b_i\right\rVert\leq\left\lVert v'_k\right\rVertvk​=a1​b1​+a2​b2​+β‹―+aiβˆ’1​biβˆ’1​+nvk′​ai​,n∈Zβˆ₯bi​βˆ₯≀βˆ₯vk′​βˆ₯
n=1n=1n=1
vkv_kvk​
LLL
βˆ₯biβˆ₯≀βˆ₯vkβˆ₯=Ξ»k(L)≀λi(L)\left\lVert b_i\right\rVert\leq\left\lVert v_k\right\rVert=\lambda_k(L)\leq\lambda_i(L)βˆ₯bi​βˆ₯≀βˆ₯vk​βˆ₯=Ξ»k​(L)≀λi​(L)
n2β‰₯4n^2\geq4n2β‰₯4
vkβ€²=p+qv_k'=p+qvk′​=p+q
ppp
vkβ€²v'_kvk′​
Liβˆ’1L_{i-1}Liβˆ’1​
βˆ₯pβˆ₯2≀βˆ₯pΒ±biβˆ₯2\left\lVert p\right\rVert^2\leq\left\lVert p\pm b_i\right\rVert^2βˆ₯pβˆ₯2≀βˆ₯pΒ±bi​βˆ₯2
βˆ₯pβˆ₯2≀14βˆ‘j=1iβˆ’1βˆ₯bjβˆ₯2\left\lVert p\right\rVert^2\leq\frac14\sum_{j=1}^{i-1}\left\lVert b_j\right\rVert^2βˆ₯pβˆ₯2≀41​j=1βˆ‘iβˆ’1​βˆ₯bj​βˆ₯2
Ξ»k2=βˆ₯vkβˆ₯2=βˆ₯a1b1+a2b2+…aiβˆ’1biβˆ’1+pβˆ₯2+n2βˆ₯qβˆ₯2\lambda_k^2=\left\lVert v_k\right\rVert^2=\left\lVert a_1b_1+a_2b_2+\dots a_{i-1}b_{i-1}+p\right\rVert^2+n^2\left\lVert q\right\rVert^2Ξ»k2​=βˆ₯vk​βˆ₯2=βˆ₯a1​b1​+a2​b2​+…aiβˆ’1​biβˆ’1​+pβˆ₯2+n2βˆ₯qβˆ₯2
βˆ₯qβˆ₯2≀14Ξ»k2\left\lVert q\right\rVert^2\leq\frac14\lambda_k^2βˆ₯qβˆ₯2≀41​λk2​
βˆ₯biβˆ₯≀14βˆ‘j=1iβˆ’1βˆ₯bjβˆ₯2+14Ξ»k2≀14βˆ‘j=1iβˆ’1max⁑(1,(54)iβˆ’4)Ξ»j(L)2+14Ξ»k(L)2≀14(1+βˆ‘j=1iβˆ’1max⁑(1,(54)iβˆ’4))Ξ»i(L)2{=max⁑(1,(54)iβˆ’4)Ξ»i(L)2iβ‰₯4<Ξ»i(L)2i=2,3\begin{align*} \left\lVert b_i\right\rVert&\leq\frac14\sum_{j=1}^{i-1}\left\lVert b_j\right\rVert^2+\frac14\lambda_k^2\\ &\leq\frac14\sum_{j=1}^{i-1}\max\left(1,\left(\frac54\right)^{i-4}\right)\lambda_j(L)^2+\frac14\lambda_k(L)^2\\ &\leq\frac14\left(1+\sum_{j=1}^{i-1}\max\left(1,\left(\frac54\right)^{i-4}\right)\right)\lambda_i(L)^2\\ &\begin{cases}=\max\left(1,\left(\frac54\right)^{i-4}\right)\lambda_i(L)^2&i\geq4\\<\lambda_i(L)^2&i=2,3\end{cases}\\ \end{align*}βˆ₯bi​βˆ₯​≀41​j=1βˆ‘iβˆ’1​βˆ₯bj​βˆ₯2+41​λk2​≀41​j=1βˆ‘iβˆ’1​max(1,(45​)iβˆ’4)Ξ»j​(L)2+41​λk​(L)2≀41​(1+j=1βˆ‘iβˆ’1​max(1,(45​)iβˆ’4))Ξ»i​(L)2{=max(1,(45​)iβˆ’4)Ξ»i​(L)2<Ξ»i​(L)2​iβ‰₯4i=2,3​​
Ξ»i(L)2≀βˆ₯biβˆ₯2\lambda_i(L)^2\leq \left\lVert b_i\right\rVert^2Ξ»i​(L)2≀βˆ₯bi​βˆ₯2
i=2,3i=2,3i=2,3
n=1n=1n=1
L=(2000002000002000002011111)L=\begin{pmatrix}2&0&0&0&0\\0&2&0&0&0\\0&0&2&0&0\\0&0&0&2&0\\1&1&1&1&1\end{pmatrix}L=​20001​02001​00201​00021​00001​​
222
bib_ibi​
βˆ₯biβˆ₯=Ξ»i\left\lVert b_i\right\rVert=\lambda_iβˆ₯bi​βˆ₯=Ξ»i​
bib_ibi​
βˆ₯biβˆ₯2=max⁑(1,(54)iβˆ’4)Ξ»i(L)2\left\lVert b_i\right\rVert^2=\max\left(1,\left(\frac54\right)^{i-4}\right)\lambda_i(L)^2βˆ₯bi​βˆ₯2=max(1,(45​)iβˆ’4)Ξ»i​(L)2